Prove that if n is a perfect square, then n +2 is not a perfect square

- lgbasallote

Prove that if n is a perfect square, then n +2 is not a perfect square

- chestercat

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- lgbasallote

i suppose first step would be to assume n is perfect square

- lgbasallote

so n = m^2

- lgbasallote

then i suppose i find n^2

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## More answers

- lgbasallote

i mean n + 2

- lgbasallote

n + 2 would be m^2 + 2

- lgbasallote

so now... i assume m^2 + 2 is a perfect square...

- anonymous

we have no 2 perfect square numbers with difference 2

- lgbasallote

so i let m^2 + 2 = k^2

- lgbasallote

@mukushla that's not really a proof.....

- shubhamsrg

you may continue @igba

- lgbasallote

hmm i suppose the next step would be
m^2 - k^2 = 2
so (m+k)(m-k) = 2
i don't think this proves anything...

- anonymous

actually that is :)
\[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

- lgbasallote

why so?

- shubhamsrg

well,,you've almost proved it..
now m and k both are integers/.

- shubhamsrg

and its k^2 -m^2 = 2 and not otherwise..

- lgbasallote

oh yes

- lgbasallote

2 = (k + m)(k-m)

- lgbasallote

now...integers huh...

- anonymous

\[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

- shubhamsrg

you know k>m..
so from
(k+m)(k-m)= 1*2
=>k+m =2
and
k-m = 1
this doesnt have an integral solution..

- lgbasallote

why do i know k > m again?

- shubhamsrg

m^2 + 2 = k^2
both m and k are positive integers
hence k>m

- shubhamsrg

in layman language,, k^2 - m^2 >0

- lgbasallote

when was it assumed that m and k are positive?

- shubhamsrg

what else do you mean by perfect square ?

- lgbasallote

i don't see how that relates.... the definition of a perfect square is x = k^2

- shubhamsrg

4 is a perfect sq since its 2^2 i.e. square of a positive integer..
and likewise..

- anonymous

@mukushla
"we have no 2 perfect square numbers with difference 2"
Because 0,1,4,9,6,5,6,9,4,1 etc....

- lgbasallote

(-2)^2 is also 4...

- shubhamsrg

you take the absolute value

- shubhamsrg

i mean you have to..

- lgbasallote

why?

- shubhamsrg

see,,
(k+m)(k-m) =2
either both are positive are both are negative,,since RHS is postive..
so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..

- lgbasallote

...i don't think this is a valid proof....

- shubhamsrg

another explanation might me x is a perfect square because its y^2
x = y^2 = (-y)^2
sqrt(x) = | y | = | -y |

- jhonyy9

who can writing here now two perfect squares with difference of 2 ?
so because i think do not exist or .... ???

- lgbasallote

since when did sqrt x become |-y|

- anonymous

0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats.
No +2 in there.

- jhonyy9

so and i think that this difference of two will be the ,,key" of this proof of ...

- jhonyy9

@lgbasallote what is your opinion from this ,please ?

- lgbasallote

my opinion is that this is some tricky algebra

- jhonyy9

how do you think it ???

- jhonyy9

so and why ?

- lgbasallote

because i can't think of a proof

- lgbasallote

how can you prove/disprove that the difference of two squares is 2...

- lgbasallote

hmmm...wait...

- jhonyy9

so i think this is very easy logicaly but to prove it will be difficile i think
so but there are again @saifoo.khan and @satellite73
probably they will can doing it
hope so much

- lgbasallote

their fields of specialty aren't exactly in proving

- anonymous

What's wrong with my proof?

- lgbasallote

anyway...
m^2 + 2 = k^2
k^2 - m^2 = 2
for it to be equal to 2...k^2 and m^2 should either be both even...or both odd
so if i assume both are even..
then k and m are also both even
so i can rewrite this as
(2x)^2 - (2y^2) = 2
4x^2 - 4y^2 = 2
then...
2x^2 - 2y^2 = 1
x^2 - y^2 = 1/2
then since it's x^2 and y^2 are both integers...the difference should also be integer
\('\therefore\) contradiction

- lgbasallote

and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

- lgbasallote

@estudier i mean

- lgbasallote

@mukushla too

- anonymous

So, that doesn't mean it's invalid..

- lgbasallote

it does actually.....you can't do infinite substitutions to prove anything....

- anonymous

There is no infinity involved, all integer squares end in the numbers I gave.

- lgbasallote

infinite loops involves infinity agree?

- anonymous

No infinite loop either, just logic.

- anonymous

An integer ends in one of digits 0 to 9.
The squares must end in etc....

- lgbasallote

but we both know that setting values is illegal in proving

- anonymous

I am not setting any values.

- lgbasallote

and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

- lgbasallote

you're checking the squares of each number

- anonymous

I am not.

- lgbasallote

and though there is a pattern...there is still susbtitution involved

- lgbasallote

and then you'll have to substitute forever to verify that the pattern really does not break

- lgbasallote

then infinity is involved

- anonymous

It's not a pattern, it's a fact
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

- lgbasallote

so you do admit that you checked the 0 to 9 squares

- anonymous

Anything else is impossible.

- lgbasallote

...that's substitution

- lgbasallote

anyway...my point is...substitution isn't really allowed in proving

- anonymous

You will find that this fact is used a lot in many valid proofs in number theory.

- lgbasallote

perhaps

- anonymous

OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

- lgbasallote

since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

- anonymous

I am not using brute force, brute force is something like testing or experimental math.
Or exhaustion (which is also a valid proof technique).
It's the use of a fact about squares (ALL squares)

- lgbasallote

can you prove it? that it's a fact about squares?

- anonymous

Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

- lgbasallote

you didn't really demonstrate anything...you just repeated what you said

- anonymous

You asked for proof, I just gave u a proof.

- anonymous

If you deny my proof, provide a counterexample.

- lgbasallote

since you try so hard to defend your proof, i'll accept it

- anonymous

That's very gracious, thank you:-)

- lgbasallote

do check on my algebra proof though please. i would like to know if that's the way contradiction works

- anonymous

OK, I will look at it now....

- anonymous

Did u do the case when both are odd?
(Perhaps a more straightforward way is one you start with n= some k squared then
next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

- lgbasallote

why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

- anonymous

No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

- lgbasallote

oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

- lgbasallote

again i ask... how is k > m

- anonymous

(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

- anonymous

factoring involves lesser steps...its obvious that k>m
\[k^2âˆ’m^2=2\]\[(kâˆ’m)(k+m)=2\]easily u can see
\[k-m=1\]\[k+m=2\]add them\[2k=3\]
Contradiction

- lgbasallote

again i ask for the nth time @mukushla how is k > m

- anonymous

ok man i just want to answer ur question

- anonymous

"2 can only be an outcome of even and even or odd and odd" is a valid statement
Why do you think it might not be valid?

- lgbasallote

well because 8 - 4 is not 2 but it's even minus even

- lgbasallote

@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

- anonymous

OK, so finally, the algebra
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

- anonymous

i thought they are positive :) sorry

- lgbasallote

why (k+1)^2 @estudier ?

- anonymous

I already, said, that's the next square....

- lgbasallote

next square?

- lgbasallote

is this still related to my proof?

- anonymous

? n is k^2, what is the next square?

- anonymous

no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) ...now add them\[2|k|=3\]Contradiction and we are done

- anonymous

No, this proof
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

- lgbasallote

are you allowed to do that? @mukushla

- lgbasallote

can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

- anonymous

why not...for any real number \[a^2=|a|^2\]

- anonymous

Given n=k^2 ->
(k+1)^2 = k^2 + 2k +1
k^2 + 2k +1 >= n +3
n +3 > n+2 QED

- anonymous

That's 2 proofs I have given now...

- lgbasallote

a few questions in mind @estudier
1) why find the next square?
2) how is k^2 + 2k + 1 > = n + 3
3) what does n + 3 > n+2 prove?

- anonymous

Think about it....:-)

- lgbasallote

in the meanwhile... @mukushla why |k| - |m| = 1?

- anonymous

because\[|k|-|m|<|k|+|m|\]

- lgbasallote

how does that make |k| - |m| = 1

- anonymous

u have 2 positive integers with ab=2 and a

**
**

- anonymous

right?

- lgbasallote

who said those are positive?

- anonymous

|k|>|m| right?

- lgbasallote

something feels wrong...

- lgbasallote

anyway...where did you get 2|k| > 3

- anonymous

ok...think on what i said...because i believe thats the shortest way to prove ur statement

- lgbasallote

|2k| = 3 *

- anonymous

adding\[|k|-|m|=1\]and\[|k|+|m|=2\]

- anonymous

Going back to your question about 2, u can just adjust it to difference of 2q...

- lgbasallote

im really skeptical about this... especially that |k|^2 - |m|^2 = 2

- lgbasallote

difference of 2q?

- anonymous

You were worried about the validity of your statement, change it to 2q instead of just 2.

- anonymous

Or 2p, or whatever letter u like best.

- anonymous

1) why find the next square?
Because we want to see that it isn't n+2
2) how is k^2 + 2k + 1 > = n + 3
Substitute n=k^2
3) what does n + 3 > n+2 prove?
See 1)

- lgbasallote

oh. when i said difference of two numbers equal to 2

- lgbasallote

but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

- lgbasallote

and @mukushla how did you get |k| > |m|

- anonymous

No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.

- lgbasallote

and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?

- anonymous

Yes, the algebra is completely general...

- lgbasallote

i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

- anonymous

...which is greater than n+3, right?

- lgbasallote

it is?

- anonymous

Number theorists don't pay attention to 0

- lgbasallote

unless k is 0....

- lgbasallote

why not?

- anonymous

Tradition, convention, call it as u like.....
So in fact there are 2 definitions for natural number , one with 0 and one without.

- anonymous

Can I try?

- lgbasallote

..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....

- anonymous

More the merrier.

- anonymous

"because i know nothing about it"
But you do now:-)

- lgbasallote

it's not like it's limited to the 99s....

- lgbasallote

i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not

- anonymous

Number theory is less onerous than real number stuff because it is integer only.
But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.

- anonymous

Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2
Now,
(m+1)^2-m^2=2m+1
Since, 2m+1>2 as m is a set of natural numbers.
Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.

- anonymous

Yes, that's similar to the one I gave...

- anonymous

"number theorists don't consider 0"
http://en.wikipedia.org/wiki/Natural_number

- anonymous

http://mathforum.org/library/drmath/view/63510.html
Using 0,1,4, etc...

- jhonyy9

ok,so who can answering this question ?
,,the line of perfect squares begin with number ... " what is this number ?

- jhonyy9

so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than
n+2=m^2 +2
suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2
so m1,2 = +/- sqrt(x^2 -2)

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